Question

In the fig, the coefficient of friction is 0.2 between the rope and the fixed pulley, and between other surface of contact, m = 0.3. Determine the minimum weight W to prevent the downward motion of the 100N body.

Answer 1

 From the given fig tanα = 3/4,

cosα = 4/5 & sinα = 3/5, 

Consider equilibrium of block W

RV = 0; R₂ = Wcosα ...(1) 

RH = 0; T₁ = µR₂ + Wsinα ...(2) 

Putting the value of equation(1) in (2)

T₁ = µWcos α + Wsinα

= 0.3 X W(4/5) + W(3/5) 

T₁ = 0.84W ...(3)

For pulley; T₂/T1 = e^µ₁^θ

T₂ = T1 X e^µ₁^θ

= 0.84We^{(0.2 X )}

T₂ = 1.574W.....(4)

Consider equilibrium of block 100N 

RV = 0; R₁ = 100cos α + R₂ ...(5) 

R₁ = 100cosα + Wcosα = 100(4/5) + W(4/5) 

R₁ = 80 + 0.8W ...(6) 

R_H = 0; 

T₂ = 100sinα –µR₁ -µR₂ 

T₂ = 100(3/5) –0.3[(80 + 0.8W) – W(4/5)] 

1.574W = 60 – 24 – 0.24W – 0.24W 

W = 17.53N.